%
\label{thm:e-index}
Let $P_k = p_1 \cdots p_k$ be the $k$th primorial
and let $n_k = p_{k+1}$ be the $(k{+}1)$th prime.
Then:
\begin{enumerate}
    \item[\textup{(i)}]
          The sequence
          $\bigl(1 + 1/n_k\bigr)^{n_k}$
          is computable in earned index arithmetic
          \textup{(I.D06, Book~I)}
          at every finite stage.

    \item[\textup{(ii)}]
          The Archimedean values satisfy
          $\bigl|\,
          (1 + 1/n_k)^{n_k} - e
          \,\bigr|
          \;\leq\;
          e / (2 n_k)$
          for all $k \geq 1$.

    \item[\textup{(iii)}]
          The primorial-weighted partial sums
          \[
              E_k
              \;:=\;
              \sum_{m=0}^{k} \frac{1}{m!}
              \;\in\; \mathbb{Q}
          \]
          satisfy $|E_k - e| \leq 3/(k{+}1)!$
          and each $E_k$ has denominator dividing~$k!$,
          hence is representable
          in the CRT tower at stage~$k$
          whenever $k! \mid P_k$.

    \item[\textup{(iv)}]
          $e$ is transcendental over~$\mathbb{Q}$
          \textup{(Hermite, 1873)}.
          In particular, $e$ is not a finite-stage
          rational;
          it arises only at the profinite limit.
\end{enumerate}

\textbf{(i)}
At stage~$k$, the index arithmetic
operates in $\mathbb{Z}/P_{k+1}\mathbb{Z}$.
The integer $n_k = p_{k+1}$
is coprime to~$P_k$,
so $1/n_k = n_k^{-1} \pmod{P_{k+1}}$ exists.
The expression $(1 + n_k^{-1})^{n_k} \pmod{P_{k+1}}$
is a well-defined integer;
its computation requires $O(\log n_k)$
modular multiplications by repeated squaring.

\textbf{(ii)}
The classical estimate for the convergence rate
of $(1+1/n)^n$ to~$e$ is
\[
    e - \Bigl(1 + \frac{1}{n}\Bigr)^n
    \;=\;
    \frac{e}{2n} + O(1/n^2).
\]
This follows from taking logarithms:
$n \log(1 + 1/n) = 1 - 1/(2n) + O(1/n^2)$,
hence $(1+1/n)^n = e^{1 - 1/(2n) + O(1/n^2)}
= e \cdot (1 - 1/(2n) + O(1/n^2))$.
Substituting $n = n_k = p_{k+1}$ gives the bound.

\textbf{(iii)}
The exponential series $e = \sum_{m=0}^\infty 1/m!$
converges absolutely.
The tail bound is
$\sum_{m=k+1}^\infty 1/m! \leq 3/(k{+}1)!$
for $k \geq 0$.
Each partial sum $E_k$ has denominator $k!$.
The condition $k! \mid P_k$ holds for~$k$
up to the largest prime in the primorial,
since $P_k = p_1 \cdots p_k$
contains all prime factors of~$k!$
when $k \leq p_k$
(which holds for all $k \geq 1$
by Bertrand's postulate).

\textbf{(iv)}
Hermite's 1873 proof of the transcendence of~$e$
is classical.
In the $\tau$-framework,
transcendence means that~$e$
does not coincide with any
finite-stage rational coordinate:
no element of
$\mathbb{Q} \subset \tau^3$
maps to~$e$ under the denotation map
(II.D23, Chapter~\ref{ch:orthodox-bridge}).
$e$ exists only as a profinite limit.